ĐKXĐ : \(xy\ne0\)
Ta có : \(\left\{{}\begin{matrix}\left(x+y\right)\left(1+\frac{1}{xy}\right)=5\\\left(x^2+y^2\right)\left(1+\frac{1}{xy}\right)=9\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left(x+y\right)\left(1+\frac{1}{xy}\right)=5\\\left(\left(x+y\right)^2-2xy\right)\left(1+\frac{1}{xy}\right)=9\end{matrix}\right.\)
- Đặt \(x+y=a,xy=b\left(b\ne0\right)\) ta được hệ là :
\(\left\{{}\begin{matrix}a\left(1+\frac{1}{b}\right)=5\\\left(a^2-2b\right)\left(1+\frac{1}{b}\right)=9\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}1+\frac{1}{b}=\frac{5}{a}\\\frac{5\left(a^2-2b\right)}{a}=9\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{1}{b}=\frac{5}{a}-1=\frac{5-a}{a}\\\frac{5\left(a^2-2b\right)}{a}=9\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}b=\frac{a}{5-a}\\\frac{5\left(a^2-\frac{2a}{5-a}\right)}{a}=9\left(I\right)\end{matrix}\right.\)
- Từ ( I ) ta có : \(5a^2-\frac{10a}{5-a}=9a\)
=> \(\frac{5a^2\left(5-a\right)}{5-a}-\frac{10a}{5-a}=\frac{9a\left(5-a\right)}{5-a}\)
=> \(5a^2\left(5-a\right)-10a=9a\left(5-a\right)\)
=> \(25a^2-5a^3-10a=45a-9a^2\)
=> \(25a^2-5a^3-10a-45a+9a^2=0\)
=> \(-5a^3+34x^2-55a=0\)
( giải phượng trình bậc 3 hen )
=> \(\left\{{}\begin{matrix}a=\frac{17+\sqrt{14}}{5}\\a=\frac{17-\sqrt{14}}{5}\\a=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}b=\frac{a}{5-a}=\frac{\frac{17+\sqrt{14}}{5}}{5-\frac{17+\sqrt{14}}{5}}=\frac{6+\sqrt{14}}{2}\\b=\frac{a}{5-a}=\frac{6-\sqrt{14}}{2}\\b=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}t^2-\frac{17+\sqrt{14}}{5}t+\frac{6+\sqrt{14}}{2}=0\\t^2-\frac{17-\sqrt{14}}{5}+\frac{6-\sqrt{14}}{2}=0\\t^2-0t+0=0\end{matrix}\right.\)
Đáp án là nghiệm của 3 phương trình trên